Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
  int m = 1;
  while (m > 0) {
    solve();
    m--;
  }
  return 0;
}
int n, l, m;
vector<vector<int> > g;
vector<vector<int> > town_people;
vector<int> c;
vector<int> tin, tout;
int timer;
vector<vector<int> > up;
vector<vector<vector<int> > > up_sets;
vector<int> _merge(vector<int> a, vector<int> b) {
  int i = 0;
  int j = 0;
  vector<int> ans;
  while ((i < a.size() || j < b.size()) && ans.size() < 10) {
    if (i == a.size()) {
      ans.push_back(b[j]);
      j++;
      continue;
    }
    if (j == b.size()) {
      ans.push_back(a[i]);
      i++;
      continue;
    }
    if (a[i] == b[j]) {
      ans.push_back(a[i]);
      i++;
      j++;
      continue;
    }
    if (a[i] > b[j]) {
      ans.push_back(b[j]);
      j++;
    } else {
      ans.push_back(a[i]);
      i++;
    }
  }
  return ans;
}
void dfs(int v, int p = 0) {
  tin[v] = ++timer;
  up[v][0] = p;
  up_sets[v][0] = town_people[v];
  for (int i = 1; i <= l; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    up_sets[v][i] = _merge(up_sets[up[v][i - 1]][i - 1], up_sets[v][i - 1]);
  }
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to != p) dfs(to, v);
  }
  tout[v] = ++timer;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = l; i >= 0; --i)
    if (!upper(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
vector<int> get_ans(int res, int a) {
  vector<int> ans;
  for (int i = l; i >= 0; --i) {
    if (!upper(up[a][i], res)) {
      ans = _merge(ans, up_sets[a][i]);
      a = up[a][i];
    }
  }
  ans = _merge(ans, up_sets[a][0]);
  return ans;
}
void solve() {
  int q;
  cin >> n >> m >> q;
  g.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int v1, v2;
    cin >> v1 >> v2;
    v1--;
    v2--;
    g[v1].push_back(v2);
    g[v2].push_back(v1);
  }
  c.resize(m);
  town_people.resize(n);
  for (int i = 0; i < m; i++) {
    scanf("%d", &c[i]);
    if (town_people[c[i] - 1].size() < 10)
      town_people[c[i] - 1].push_back(i + 1);
  }
  tin.resize(n), tout.resize(n), up.resize(n), up_sets.resize(n);
  l = 1;
  while ((1 << l) <= n) ++l;
  for (int i = 0; i < n; ++i) {
    up[i].resize(l + 1);
    up_sets[i].resize(l + 1);
  }
  dfs(0);
  for (int i = 0; i < q; i++) {
    int u, v, t;
    scanf("%d%d%d", &u, &v, &t);
    u--;
    v--;
    int res = lca(u, v);
    vector<int> r1 = get_ans(res, u);
    vector<int> r2 = get_ans(res, v);
    vector<int> ans = _merge(r1, r2);
    ans = _merge(ans, town_people[res]);
    if (t >= ans.size()) {
      printf("%d ", ans.size());
      for (int j = 0; j < ans.size(); j++) {
        printf("%d ", ans[j]);
      }
      printf("\n");
    } else {
      printf("%d ", t);
      for (int j = 0; j < t; j++) {
        printf("%d ", ans[j]);
      }
      printf("\n");
    }
  }
}